[tex]\bf f(x)= 5x^3-7x^2+7\implies \cfrac{dy}{dx}=15x^2-14x
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\textit{what's the slope when x = 3?}
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\cfrac{dy}{dx}=15(3)^2-14(3)\implies \cfrac{dy}{dx}=93
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\textit{so, now we know that when x =3 }
\begin{cases}
x=3\\
y=79\\
m=93
\end{cases}
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\textit{using the point-slope form then}
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y-79=93(x-3)\implies y=93x-200[/tex]
